A car traveling at a constant speed of 33.4 m/s passes a trooper hidden behind a billboard. One second later the trooper starts the car with a constant acceleration of 3.28 m/s2 How long after the trooper starts the chase does he overtake the speeding car
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To solve this problem, we can use the kinematic equations of motion.
Let's denote the time it takes for the trooper to overtake the speeding car as "t".
For the speeding car:
Initial velocity, u = 33.4 m/s (constant speed)
Acceleration, a = 0 m/s^2 (since the car is traveling at a constant speed)
For the trooper's car:
Initial velocity, u = 0 m/s (since the car starts from rest)
Acceleration, a = 3.28 m/s^2 (constant acceleration)
We need to find the time at which the positions (or distances traveled) of the two cars are equal.
The equation to use is:
Distance = Initial Velocity × Time + 0.5 × Acceleration × Time^2
Distance for the speeding car, D1 = 33.4 × t
Distance for the trooper's car, D2 = 0.5 × 3.28 × t^2
Since they are equal at the point of overtaking, we can equate the distances:
33.4 × t = 0.5 × 3.28 × t^2
Now, let's solve this equation to find the value of "t".
First, divide both sides of the equation by t:
33.4 = 0.5 × 3.28 × t
Next, divide both sides of the equation by (0.5 × 3.28):
33.4 / (0.5 × 3.28) = t
Now, calculate the value of t by evaluating the right-hand side of the equation:
t ≈ 10.18 seconds
Therefore, the trooper will overtake the speeding car approximately 10.18 seconds after starting the chase.